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FrogsWithSock
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Homework Statement
Titrating .1006M CH3COOH with .2012M NaOH. pKa=4.74, Ka=1.8*10^-5
What is the pH at the equivalence point of the titration?
Homework Equations
pKa = -log(Ka) pH = -log[H+]
The Attempt at a Solution
Alright, after setting up an ICE table (in order to find [H+] at the equivalence point) I realized that when you work it out Ka = [H+], however when I plugged that in I got a pH in th 4-5 range when it should be above 7.
Here is my Ka reasoning
Ka = [A-][H+]/[HA]
Ka = [Co+x][x]/[Co-x] Co = Initial Concentration
x is very small (Weak acid, no/little disassociation)
Ka = x
Ka = [H+]
Am I missing something or is my logic just wrong? Anything that could point me in the correct direction would be greatly appreciated.